Let $h(x)=-6\sin(x)-3x$. $h'(x)=$
The expression for $h(x)$ includes $\sin(x)$. Remember that the derivative of $\sin(x)$ is $\cos(x)$. Put another way, $\dfrac{d}{dx}[\sin(x)]=\cos(x)$. $\begin{aligned} h'(x)&=\dfrac{d}{dx}[-6\sin(x)-3x] \\\\ &=-6\dfrac{d}{dx}[\sin(x)]-3\dfrac{d}{dx}(x) \\\\ &=-6\cdot\cos(x)-3\cdot1 \\\\ &=-6\cos(x)-3 \end{aligned}$ In conclusion, $h'(x)=-6\cos(x)-3$